Calculus II Final Exam Practice

Problem 1 · Integration by Parts

Evaluate \(\int_1^e x^2\ln x\,dx\).

Solution. Choose \(u=\ln x\) and \(dv=x^2\,dx\), so \(du=\frac{dx}{x}\) and \(v=\frac{x^3}{3}\). The logarithm is differentiated because its derivative is simpler, and the polynomial is integrated. Then

Combining over a common denominator gives \(\frac{3e^3-e^3+1}{9}\).

Answer: \(\dfrac{2e^3+1}{9}\approx 4.57\).

Problem 2 · A Cyclic Integral

Find \(\int e^{2x}\cos 3x\,dx\).

Solution. Integrate by parts twice and solve for the integral. Let \(I=\int e^{2x}\cos3x\,dx\). With \(u=\cos3x\) and \(dv=e^{2x}dx\), we get \(I=\frac12e^{2x}\cos3x+\frac32\int e^{2x}\sin3x\,dx\). Applying parts again to the last integral, with \(u=\sin3x\), gives \(\int e^{2x}\sin3x\,dx=\frac12e^{2x}\sin3x-\frac32I\). Substituting,

Moving \(\frac94I\) to the left gives \(\frac{13}{4}I=\frac{e^{2x}}{4}(2\cos3x+3\sin3x)\), so \(I\) is the expression below.

Answer: \(\dfrac{e^{2x}\,(2\cos3x+3\sin3x)}{13}+C\). Differentiating confirms it.

Problem 3 · Substitution Before Parts

Evaluate \(\int x^3e^{x^2}\,dx\).

Solution. Split \(x^3=x^2\cdot x\) and let \(w=x^2\), so \(dw=2x\,dx\). The integral becomes \(\frac12\int we^{w}\,dw\), which is a single integration by parts: \(\int we^w\,dw=(w-1)e^w\). Therefore

Answer: \(\tfrac12(x^2-1)e^{x^2}+C\).

Problem 4 · Powers of Sine and Cosine

Find \(\int\sin^3x\cos^2x\,dx\).

Solution. The power of sine is odd, so save one factor \(\sin x\,dx\) for \(-du\) with \(u=\cos x\) and convert the remaining \(\sin^2x\) with \(\sin^2x=1-\cos^2x\):

Answer: \(-\dfrac{\cos^3x}{3}+\dfrac{\cos^5x}{5}+C\).

Problem 5 · Powers of Tangent

Evaluate \(\int_0^{\pi/4}\tan^3x\,dx\).

Solution. Write \(\tan^3x=\tan x(\sec^2x-1)\). Then \(\int\tan x\sec^2x\,dx=\frac{\tan^2x}{2}\) (let \(u=\tan x\)), and \(\int\tan x\,dx=-\ln|\cos x|\). So

Answer: \(\dfrac12-\dfrac12\ln2\approx0.153\).

Problem 6 · Trigonometric Substitution

Evaluate \(\int_0^2\dfrac{x^2}{\sqrt{4-x^2}}\,dx\).

Solution. The radical \(\sqrt{a^2-x^2}\) with \(a=2\) suggests \(x=2\sin\theta\), so \(dx=2\cos\theta\,d\theta\) and \(\sqrt{4-x^2}=2\cos\theta\). The limits \(x=0\) and \(x=2\) become \(\theta=0\) and \(\theta=\frac\pi2\). The integral becomes

Answer: \(\pi\).

Problem 7 · Partial Fractions

Find \(\int\dfrac{3x+5}{(x-1)(x+3)}\,dx\).

Solution. Write \(\frac{3x+5}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}\). Multiplying out gives \(3x+5=A(x+3)+B(x-1)\). Setting \(x=1\) gives \(8=4A\), so \(A=2\). Setting \(x=-3\) gives \(-4=-4B\), so \(B=1\). Hence

Answer: \(2\ln|x-1|+\ln|x+3|+C\).

Problem 8 · An Irreducible Quadratic Factor

Find \(\int\dfrac{2x^2+3}{x(x^2+1)}\,dx\) for \(x>0\).

Solution. Set \(\frac{2x^2+3}{x(x^2+1)}=\frac Ax+\frac{Bx+C}{x^2+1}\). Then \(2x^2+3=A(x^2+1)+(Bx+C)x\). At \(x=0\), \(A=3\). Comparing the \(x^2\) coefficients, \(2=A+B\), so \(B=-1\); comparing the \(x\) coefficients, \(C=0\). Thus the integrand is \(\frac3x-\frac{x}{x^2+1}\), and

Answer: \(3\ln x-\tfrac12\ln(x^2+1)+C\).

Problem 9 · Completing the Square

Find \(\int\dfrac{dx}{x^2+4x+13}\).

Solution. The discriminant is \(16-52<0\), so the quadratic has no real roots. Complete the square: \(x^2+4x+13=(x+2)^2+9\). With \(u=x+2\), the integral is \(\int\frac{du}{u^2+9}=\frac13\arctan\frac u3\). Therefore

Answer: \(\dfrac13\arctan\dfrac{x+2}{3}+C\).

Problem 10 · An Infinite Interval

Evaluate \(\int_1^\infty\dfrac{dx}{x(x+1)}\).

Solution. Decompose \(\frac{1}{x(x+1)}=\frac1x-\frac{1}{x+1}\). Then for \(b>1\),

As \(b\to\infty\), \(\frac{b}{b+1}\to1\), so \(\ln\frac{b}{b+1}\to0\).

Answer: \(\ln2\). The integral converges.

Problem 11 · A Singular Integrand

Evaluate \(\int_0^1\ln x\,dx\), or show that it diverges.

Solution. The integrand is unbounded at \(x=0\), so write the integral as \(\lim_{t\to0^+}\int_t^1\ln x\,dx\). An antiderivative is \(x\ln x-x\), by parts. Then

As \(t\to0^+\), \(t\ln t\to0\) (rewrite as \(\frac{\ln t}{1/t}\) and apply L'Hôpital's rule), and \(t\to0\).

Answer: \(-1\). The integral converges.

Problem 12 · Convergence by Comparison

Determine whether \(\int_2^\infty\dfrac{dx}{\sqrt{x^3-1}}\) converges.

Solution. For \(x\ge2\), we have \(1\le\frac{x^3}{8}\), so \(x^3-1\ge x^3-\frac{x^3}{8}=\frac78x^3\). Hence \(\sqrt{x^3-1}\ge\sqrt{\tfrac78}\,x^{3/2}\) and

The integral \(\int_2^\infty x^{-3/2}\,dx\) converges because \(p=\frac32>1\). By the direct comparison test, the given integral converges too.

Answer: It converges.

Problem 13 · Cylindrical Shells

The region between \(y=x\) and \(y=x^2\), for \(0\le x\le1\), is rotated about the \(y\)-axis. Find the volume.

Solution. Use shells. A strip at distance \(x\) from the axis has height \(x-x^2\) and sweeps a cylinder of radius \(x\):

Answer: \(\dfrac{\pi}{6}\).

Problem 14 · Arc Length

Find the length of \(y=\frac23x^{3/2}\) for \(0\le x\le3\).

Solution. Here \(y'=x^{1/2}\), so \(1+(y')^2=1+x\). The arc length is

Answer: \(\dfrac{14}{3}\).

Problem 15 · Work

A spring has natural length \(1\) m, and \(12\) J of work stretches it from length \(1\) m to length \(2\) m. How much work is needed to stretch it from \(2\) m to \(3\) m?

Solution. Hooke's law gives a force \(F=ks\) when the spring is stretched \(s\) meters beyond its natural length. The first stretch takes work \(\int_0^1ks\,ds=\frac k2=12\), so \(k=24\) N/m. The second stretch goes from \(s=1\) to \(s=2\):

Answer: \(36\) J.

Problem 16 · A Sequence Limit

Find \(\lim_{n\to\infty}\bigl(1+\frac2n\bigr)^n\).

Solution. This is an indeterminate form \(1^\infty\). Let \(L\) be the limit; then \(\ln L=\lim n\ln\bigl(1+\frac2n\bigr)\). Substituting \(h=\frac1n\), this is \(\lim_{h\to0^+}\frac{\ln(1+2h)}{h}=2\) by L'Hôpital's rule (the derivative of the numerator is \(\frac{2}{1+2h}\to2\)). Therefore \(\ln L=2\).

Answer: \(e^2\approx7.389\).

Problem 17 · A Recursive Sequence

Let \(a_1=\sqrt2\) and \(a_{n+1}=\sqrt{2+a_n}\). Show that \((a_n)\) converges and find its limit.

Solution. We show by induction that \(a_n<2\) and \(a_n<a_{n+1}\). First, \(a_1=\sqrt2<2\). If \(a_n<2\), then \(a_{n+1}=\sqrt{2+a_n}<\sqrt4=2\). For monotonicity, \(a_{n+1}^2-a_n^2=2+a_n-a_n^2=(2-a_n)(1+a_n)>0\) whenever \(0<a_n<2\), so \(a_{n+1}>a_n\). A bounded increasing sequence converges, by the Monotone Convergence Theorem. If the limit is \(L\), then \(L=\sqrt{2+L}\), so \(L^2-L-2=0\), giving \((L-2)(L+1)=0\). Since \(L>0\),

Answer: The sequence converges to \(L=2\).

Problem 18 · A Telescoping Series

Find \(\sum_{n=1}^\infty\dfrac{1}{n(n+2)}\).

Solution. Decompose \(\frac{1}{n(n+2)}=\frac12\bigl(\frac1n-\frac1{n+2}\bigr)\). The partial sum telescopes, leaving only the first two positive terms and the last two negative terms:

Answer: \(\dfrac34\).

Problem 19 · The Integral Test

Does \(\sum_{n=2}^\infty\dfrac{1}{n(\ln n)^2}\) converge?

Solution. The function \(f(x)=\frac{1}{x(\ln x)^2}\) is positive, continuous, and decreasing for \(x\ge2\). With \(u=\ln x\),

The integral is finite, so the series converges by the integral test.

Answer: It converges. (The integral equals \(1/\ln2\approx1.44\), which bounds the tail but is not the sum of the series.)

Problem 20 · The Alternating Series Test

Classify \(\sum_{n=1}^\infty\dfrac{(-1)^n}{\sqrt n}\) as absolutely convergent, conditionally convergent, or divergent.

Solution. The terms \(b_n=\frac{1}{\sqrt n}\) are positive, decreasing, and tend to \(0\), so the series converges by the alternating series test. The series of absolute values is \(\sum n^{-1/2}\), a \(p\)-series with \(p=\frac12\le1\), which diverges.

Answer: The series converges conditionally.

Problem 21 · The Ratio Test and a Sum

Show that \(\sum_{n=1}^\infty\dfrac{n^2}{3^n}\) converges and find its sum.

Solution. The ratio test gives \(\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{3n^2}\to\frac13<1\), so the series converges. For the sum, use the known power series \(\sum_{n=1}^\infty n^2x^n=\frac{x(1+x)}{(1-x)^3}\) for \(|x|<1\), which comes from differentiating \(\sum x^n=\frac{1}{1-x}\) twice. At \(x=\frac13\),

Answer: The series converges, and its sum is \(\dfrac32\).

Problem 22 · Radius and Interval of Convergence

Find the radius and the interval of convergence of \(\sum_{n=1}^\infty\dfrac{(x-2)^n}{n\,3^n}\).

Solution. By the ratio test, \(\bigl|\frac{a_{n+1}}{a_n}\bigr|=\frac{|x-2|}{3}\cdot\frac{n}{n+1}\to\frac{|x-2|}{3}\). The series converges when \(|x-2|<3\), so \(R=3\) and the open interval is \((-1,5)\). At \(x=5\) the series is \(\sum\frac1n\), the harmonic series, which diverges. At \(x=-1\) it is \(\sum\frac{(-1)^n}{n}\), which converges by the alternating series test.

Answer: \(R=3\), and the interval of convergence is \([-1,5)\).

Problem 23 · Differentiating a Series

Use \(\sum_{n=0}^\infty x^n=\frac{1}{1-x}\) to find \(\sum_{n=1}^\infty\dfrac{n}{2^n}\).

Solution. Differentiate both sides for \(|x|<1\): \(\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\). Multiplying by \(x\) gives \(\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\). Now set \(x=\frac12\), which lies inside the interval of convergence:

Answer: \(2\).

Problem 24 · A Series for the Arctangent

Starting from \(\frac{1}{1+t^2}=\sum_{n=0}^\infty(-1)^nt^{2n}\), find a series for \(\arctan x\) and use it to approximate \(\arctan(0.1)\) with error less than \(10^{-5}\).

Solution. Integrate term by term from \(0\) to \(x\) for \(|x|<1\):

At \(x=0.1\) the series alternates with decreasing terms, so the error after a partial sum is less than the first omitted term. Using \(x-\frac{x^3}{3}=0.1-0.000333\ldots\), the first omitted term is \(\frac{(0.1)^5}{5}=2\times10^{-6}<10^{-5}\).

Answer: \(\arctan(0.1)\approx0.099667\), correct to within \(2\times10^{-6}\).

Problem 25 · A Taylor Polynomial

Find the degree-2 Taylor polynomial of \(f(x)=\sqrt x\) about \(x=4\), and use it to approximate \(\sqrt{4.4}\).

Solution. We have \(f(4)=2\), \(f'(x)=\frac{1}{2\sqrt x}\) so \(f'(4)=\frac14\), and \(f''(x)=-\frac{1}{4}x^{-3/2}\) so \(f''(4)=-\frac{1}{32}\). The polynomial is

At \(x=4.4\): \(2+0.1-\frac{0.16}{64}=2+0.1-0.0025=2.0975\).

Answer: \(T_2(x)=2+\tfrac14(x-4)-\tfrac1{64}(x-4)^2\), and \(\sqrt{4.4}\approx2.0975\) (the true value is \(2.09762\ldots\)).

Problem 26 · A Limit by Series

Find \(\lim_{x\to0}\dfrac{\sin x-x+\frac{x^3}{6}}{x^5}\).

Solution. Use the Maclaurin series \(\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots\). Then the numerator is \(\frac{x^5}{120}-\frac{x^7}{5040}+\cdots\), and dividing by \(x^5\) gives \(\frac1{120}-\frac{x^2}{5040}+\cdots\). As \(x\to0\), only the constant survives.

Answer: \(\dfrac1{120}\).

Problem 27 · Lagrange Error Bound

Approximate \(\sin(0.5)\) by the degree-3 Taylor polynomial about \(0\), and bound the error.

Solution. The degree-3 polynomial is \(T_3(x)=x-\frac{x^3}{6}\), so \(T_3(0.5)=0.5-\frac{0.125}{6}=0.479167\). The Lagrange remainder is \(R_3(x)=\frac{f^{(4)}(c)}{4!}x^4\) for some \(c\) between \(0\) and \(x\), and \(|f^{(4)}(c)|=|\sin c|\le1\). This gives \(|R_3(0.5)|\le\frac{(0.5)^4}{24}=\frac{1}{384}\). A sharper bound is available because the \(x^4\) coefficient of sine is zero, so \(T_3=T_4\) and we may use the remainder \(R_4(x)=\frac{f^{(5)}(c)}{5!}x^5\), where \(|f^{(5)}(c)|=|\cos c|\le1\): \(|R_4(0.5)|\le\frac{(0.5)^5}{120}=\frac{1}{3840}\).

Answer: \(\sin(0.5)\approx0.479167\), with error at most \(\frac{1}{3840}\approx2.6\times10^{-4}\). The true value is \(0.479426\), and the actual error \(2.59\times10^{-4}\) is within the bound.

Problem 28 · Parametric Derivatives

For \(x=t^2\), \(y=t^3-3t\), find \(\frac{dy}{dx}\) at \(t=2\) and locate the points with a horizontal tangent.

Solution. Compute \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2-3\). Then

A horizontal tangent needs \(\frac{dy}{dt}=0\) with \(\frac{dx}{dt}\ne0\): \(3t^2-3=0\) gives \(t=\pm1\), and \(2t\ne0\) at both. The points are \((1,-2)\) at \(t=1\) and \((1,2)\) at \(t=-1\).

Answer: \(\frac{dy}{dx}=\frac94\) at \(t=2\); horizontal tangents at \((1,-2)\) and \((1,2)\).

Problem 29 · Parametric Arc Length

Find the length of the curve \(x=t^2\), \(y=\frac23t^3\) for \(0\le t\le1\).

Solution. Here \(x'=2t\) and \(y'=2t^2\), so \(\sqrt{(x')^2+(y')^2}=\sqrt{4t^2+4t^4}=2t\sqrt{1+t^2}\) for \(t\ge0\). With \(u=1+t^2\),

Answer: \(\dfrac{2(2\sqrt2-1)}{3}\approx1.219\).

Problem 30 · Polar Area

Find the area enclosed by the cardioid \(r=2+2\cos\theta\).

Solution. The curve is traced once for \(0\le\theta\le2\pi\), and the area formula is \(A=\frac12\int r^2\,d\theta\). Expand \(r^2=4+8\cos\theta+4\cos^2\theta\) and use \(\cos^2\theta=\frac{1+\cos2\theta}{2}\):

Answer: \(6\pi\approx18.85\).